If cot θ = $\dfrac{q}{p}$, show that $\Big(\dfrac{p\sin θ - q\cos θ}{p\sin θ+ q\cos θ}\Big)= \dfrac{p^2 - q^2}{p^2 + q^2}$.

cot θ = $\dfrac{\text{base}}{\text{perpendicular}} = \dfrac{q}{p}$

Let base = qx and perpendicular = px

We will find hypotenuse by using pythagoras theorem

Hypotenuse² = Base² + Perpendicular²

Hypotenuse² = (qx)² + (px)²

Hypotenuse² = (q² + p²)x²

Hypotenuse = $\sqrt{(q^2 + p^2)x^2}$

Hypotenuse = $\sqrt{(q^2 + p^2)}x$

Now,

sin θ = $\dfrac{\text{perpendicular}}{\text{hypotenuse}}= \dfrac{px}{\sqrt{(p^2 + q^2)}x} = \dfrac{p}{\sqrt{p^2 + q^2}}$

cos θ = $\dfrac{\text{base}}{\text{hypotenuse}}= \dfrac{qx}{\sqrt{(p^2 + q^2)}x} = \dfrac{q}{\sqrt{p^2 + q^2}}$

Substituting values we get :

$\Rightarrow \dfrac{p\sin θ - q\cos θ}{p\sin θ+ q\cos θ}\\[1em] = \dfrac{p\times\dfrac{p}{\sqrt{p^2 + q^2}} - q\times\dfrac{q}{\sqrt{p^2 + q^2}}}{p\times\dfrac{p}{\sqrt{p^2 + q^2}} + q\times\dfrac{q}{\sqrt{p^2 + q^2}} }\\[1em] = \dfrac{\dfrac{p^2}{\sqrt{p^2 + q^2}} - \dfrac{q^2}{\sqrt{p^2 + q^2}} }{\dfrac{p^2}{\sqrt{p^2 + q^2}} + \dfrac{q^2}{\sqrt{p^2 + q^2}}}\\[1em] = \dfrac{\dfrac{p^2 - q^2}{\sqrt{p^2 + q^2}}}{\dfrac{p^2 + q^2}{\sqrt{p^2 + q^2}} }\\[1em] = \dfrac{(p^2 - q^2)\times {\sqrt{p^2 + q^2}} }{(p^2 + q^2)\times {\sqrt{p^2 + q^2}} }\\[1em] = \dfrac{p^2 - q^2}{p^2 + q^2}.$

Hence, proved that $\dfrac{p\sin θ - q\cos θ}{p\sin θ+ q\cos θ}$ = $\dfrac{p^2 - q^2}{p^2 + q^2}$.